zoukankan      html  css  js  c++  java
  • HDU-1087 Super Jumping! Jumping! Jumping! (DP)

    Problem Description

    Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.

    这里写图片描述

    The game can be played by two or more than two players. It consists of a chessboard(棋盘)and some chessmen(棋子), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
    Your task is to output the maximum value according to the given chessmen list.

    Input

    Input contains multiple test cases. Each test case is described in a line as follow:
    N value_1 value_2 …value_N
    It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
    A test case starting with 0 terminates the input and this test case is not to be processed.

    Output

    For each case, print the maximum according to rules, and one line one case.

    Sample Input

    3 1 3 2
    4 1 2 3 4
    4 3 3 2 1
    0

    Sample Output

    4
    10
    3


    思路:

    类似求最长上升子序列O(n^2)的做法,求的是最大和数组内存的改成以i结尾的上升序列和。

    Code:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define MAXN 100000+10
     4 int a[1000], dp[MAXN];
     5 
     6 int main()
     7 {
     8     int n;
     9     while(scanf("%d", &n), n){
    10         for (int i = 1; i <= n; ++i){
    11             scanf("%d", &a[i]);
    12             dp[i] = a[i];
    13         }
    14         for (int i = 2; i <= n; ++i)
    15             for (int j = 1; j < i; ++j)
    16                 if (a[j] < a[i])
    17                     dp[i] = max(dp[i], dp[j]+a[i]);
    18         int ans = dp[1];
    19         for (int i = 2; i <= n; ++i)
    20             ans = max(ans, dp[i]);
    21         printf("%d
    ", ans);
    22     }
    23 
    24     return 0;
    25 }
  • 相关阅读:
    Swift技术之如何在iOS 8下使用Swift设计一个自定义的输入法 (主要是NSLayoutConstraint 的使用)
    android 旋转手机的时候,如何忽略onCreate再次被系统调用?
    在iOS 8中使用UIAlertController
    09_android入门_采用android-async-http开源项目的GET方式或POST方式实现登陆案例
    一些工具的版本问题 valgrind gdb 以及编译
    C struct __attribute__ ((__packed__))
    C++ class 只允许堆创建/只允许栈创建
    Shell 字符串操作
    存储系统的分类
    ssh 到服务器然后输入中文保存到本地变成乱码
  • 原文地址:https://www.cnblogs.com/robin1998/p/6359120.html
Copyright © 2011-2022 走看看